Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > ans, res;
int bas[30 + 2], sbas[30 + 2];
int pos[10010], x[10010], y[10010], sy[10010];
int insert(int p) {
  int s = 0;
  for (int i = 30; i >= 0; i--)
    if ((x[p] >> i) & 1) {
      if (bas[i])
        x[p] ^= x[bas[i]], s ^= sbas[i];
      else {
        bas[i] = p;
        sbas[i] = s | (1 << i);
        return i;
      }
    }
  return -1;
}
bool find(int &y, int &s) {
  for (int i = 30; i >= 0; i--)
    if ((y >> i) & 1) {
      if (bas[i])
        y ^= x[bas[i]], s ^= sbas[i];
      else
        return false;
    }
  return true;
}
int q[10010];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &x[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &y[i]);
  for (int i = 1; i <= n; i++) pos[i] = insert(i);
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    int s = 0;
    if (!find(y[i], s)) {
      puts("-1");
      return 0;
    }
    if (pos[i] >= 0) {
      q[cnt] = i;
      sy[cnt++] = s;
      continue;
    }
    ans.push_back(make_pair(i, i));
    for (int j = 30; j >= 0; j--)
      if ((s >> j) & 1) ans.push_back(make_pair(i, bas[j]));
  }
  for (int i = 0; i < cnt; i++) {
    int s = sy[i];
    sy[i] = 0;
    for (int j = 0; j < cnt; j++)
      if ((s >> pos[q[j]]) & 1) sy[i] |= 1 << j;
  }
  for (int i = 0; i < cnt; i++) {
    if (!((sy[i] >> i) & 1)) {
      for (int j = i + 1; j < cnt; j++)
        if ((sy[j] >> i) & 1) {
          res.push_back(make_pair(q[i], q[j]));
          res.push_back(make_pair(q[j], q[i]));
          res.push_back(make_pair(q[i], q[j]));
          swap(sy[i], sy[j]);
          break;
        }
    }
    if ((sy[i] >> i) & 1)
      for (int j = i + 1; j < cnt; j++)
        if ((sy[j] >> i) & 1)
          res.push_back(make_pair(q[j], q[i])), sy[j] ^= sy[i];
  }
  reverse(res.begin(), res.end());
  for (int i = 0; i < cnt; i++) {
    if (!((sy[i] >> i) & 1)) ans.push_back(make_pair(q[i], q[i]));
    for (int j = i + 1; j < cnt; j++)
      if ((sy[i] >> j) & 1) ans.push_back(make_pair(q[i], q[j]));
  }
  for (pair<int, int> i : res) ans.push_back(i);
  printf("%d\n", ans.size());
  for (pair<int, int> i : ans) printf("%d %d\n", i.first, i.second);
  return 0;
}